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Q.

For a first order reaction when log(a – x) on y-axis is plotted against time on x-axis. Slope of the graph will be equal to('a' is the initial concentration of the reactant and ‘x’ is concentration of reactant consumed in time ‘t’ )

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a

$- \frac{K}{2 .303}$

b

$\frac{0 .693}{10}$

c

$K$

d

$\frac{2 .303}{K}$

answer is A.

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Detailed Solution

$K = \frac{2 .303}{t} \log \frac{a}{a - x} loga - \log (a - x) = \frac{K}{2 .303} \times t \log (a - x) = loga - \frac{K}{2 .303} \times t \underset{y}{\log (a - x)} = \underset{c}{loga} - \underset{m}{\frac{K}{2 .303}} \times \underset{x}{t} Slope = - \frac{K}{2 .303}$

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For a first order reaction when log(a – x) on y-axis is plotted against time on x-axis. Slope of the graph will be equal to('a' is the initial concentration of the reactant and ‘x’ is concentration of reactant consumed in time ‘t’ )